Problem 12 - Calculus Early Transcendentals

Question

Find the vertices and foci of the ellipse and sketch its graph. $$\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$$

Step-by-Step Solution

Answer

Vertices: (0, 10), (0, -10); Foci: (0, 6), (0, -6).

Step 1: Identify the standard form of the ellipse equation

The given equation is \(\frac{x^{2}}{64} + \frac{y^{2}}{100} = 1\). This corresponds to the standard form of an ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \(a^2 = 64\) and \(b^2 = 100\). Since \(b^2 > a^2\), this ellipse is vertical.

Step 2: Find lengths of semi-major and semi-minor axes

Calculate \(a\) and \(b\) from \(a^2\) and \(b^2\): \(a = \sqrt{64} = 8\) and \(b = \sqrt{100} = 10\). Since \(b > a\), \(b\) is the semi-major axis length and \(a\) is the semi-minor axis length.

Step 3: Determine vertices of the ellipse

Vertices for a vertical ellipse are located on the y-axis at \((0, \pm b)\). Therefore, the vertices are \((0, 10)\) and \((0, -10)\).

Step 4: Calculate the distance from the center to the foci

Use the formula for the foci distance \(c\) in an ellipse, \(c = \sqrt{b^2 - a^2}\). Substituting the known values, \(c = \sqrt{100 - 64} = \sqrt{36} = 6\).

Step 5: Locate the foci of the ellipse

Since the ellipse is vertical, the foci are located along the y-axis at \((0, \pm c)\). Therefore, the foci are \((0, 6)\) and \((0, -6)\).

Step 6: Sketch the graph of the ellipse

The center of the ellipse is at the origin \((0, 0)\). The major axis length is 20 (10 in each direction along the y-axis) and the minor axis length is 16 (8 in each direction along the x-axis). Plot these points and draw an ellipse through them, ensuring the ellipse is taller than it is wide.

Key Concepts

Standard Form of Ellipse

Understanding the equation of an ellipse is crucial for analyzing its properties. An ellipse is typically represented in its standard form, which is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This formula involves two principal parameters, \(a\) and \(b\), which are the denominators of the \(x^2\) and \(y^2\) terms, respectively. These parameters dictate the shape and orientation of the ellipse. In the given exercise, the equation is \(\frac{x^2}{64} + \frac{y^2}{100} = 1\). By comparing it to the standard form, we identify \(a^2 = 64\) and \(b^2 = 100\). Since \(b^2\) is greater than \(a^2\), it indicates that the ellipse is elongated along the y-axis, making it a vertical ellipse. This is an important distinction as it determines how the vertices and foci will be aligned.

Vertices of Ellipse

Vertices in an ellipse are the endpoints of the major axis. For a vertical ellipse (where \(b > a\)), the vertices lie on the y-axis. The standard positions of vertices are given by \((0, \pm b)\). Probability wise, this means we move up and down from the center, located at the origin \((0, 0)\) in this case, by the semi-major axis length \(b\). From our calculations, since \(b = \sqrt{100} = 10\), the vertices of the given ellipse are \((0, 10)\) and \((0, -10)\). This positioning defines how high and low the ellipse reaches along the y-axis, establishing the vertical nature of the curve.

Foci of Ellipse

The foci are two fixed points inside the ellipse that are crucial for its definition. The sum of the distances from any point on the ellipse to the two foci is constant. To find the foci, we first need to calculate the value of \(c\), which represents the distance from the center to each focus. For any ellipse, this is calculated using \(c = \sqrt{b^2 - a^2}\). In our scenario, substituting \(b^2 = 100\) and \(a^2 = 64\), we get \(c = \sqrt{100 - 64} = \sqrt{36} = 6\). Since this ellipse is vertical, the foci will be located along the y-axis at positions \((0, \pm c)\) from the center. Hence, the foci are at \((0, 6)\) and \((0, -6)\). These points further emphasize the vertical elongation of the ellipse.

Semi-Major and Semi-Minor Axes

In an ellipse, the **semi-major axis** and the **semi-minor axis** are critical components that define its size. The **semi-major axis** is the longest radius of the ellipse, extending from the center to the ellipse's farthest point. Conversely, the **semi-minor axis** is the shortest radius. In the standard form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), if \(b > a\), \(b\) serves as the semi-major axis and \(a\) as the semi-minor axis. From our values, \(b = 10\) and \(a = 8\). Therefore, we conclude that the semi-major axis is 10 units and the semi-minor axis is 8 units. These measurements determine how the ellipse stretches and compresses in its structure, making the ellipse taller than it is wide in this instance.